Chapter 9, Problem 43P

(a)

To determine

The velocity of the third particle.

Answer to Problem 43P

The velocity of third particle is $\left(-9.33\times {10}^6\widehat{i}-8.33\times {10}^6\widehat{j}\right)\text{m/s}$ .

Explanation of Solution

Atomic nucleous splits into three particles hence their momentum remains conserved as no extrenal force is applied.

Write the expession for the conservation of the momentum.

  $m_n\overrightarrow{v_n}=m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+m_3\overrightarrow{v_3}$                                                                            (I)

Here $m_n$ is the mass of the nucleus, $\overrightarrow{v_n}$ is the velocity of nucleus, $m_1$ is the mass of first paricle, $\overrightarrow{v_1}$ is the velocit of first particle, $m_2$ is the mass of second particle, $\overrightarrow{v_2}$ is the velocity of second particle, $m_3$ is the mass of the third particle and $\overrightarrow{v_3}$ is the velocity of the third particle.

As nucleus is at rest. Substitute $0\text{m/s}$ for $\overrightarrow{v_n}$ and simplify equation (I) for  $\overrightarrow{v_3}$.

    $\overrightarrow{v_3}=\frac{-\left(m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}\right)}{m_3}$                                                                                    (II)

Conclusion:

Substitute $5\times {10}^{-27}\text{kg}$ for $m_1$, $\left(6\times {10}^6\text{m/s}\right)\widehat{j}$ for $\overrightarrow{v_1}$, $8.4\times {10}^{-27}\text{kg}$ for $m_2$, $\left(4\times {10}^6\text{m/s}\right)\widehat{i}$ for $\overrightarrow{v_2}$ and $3.6\times {10}^{-27}\text{kg}$ for $m_3$ in equation (II).

    $\begin{array}{c}\overrightarrow{v_3}=\frac{-\left[\left(5\times {10}^{-27}\text{kg}\right)\left(\left(6\times {10}^6\text{m/s}\right)\widehat{j}\right)+\left(8.4\times {10}^{-27}\text{kg}\right)\left(\left(4\times {10}^6\text{m/s}\right)\widehat{i}\right)\right]}{3.6\times {10}^{-27}\text{kg}}\\ =\left(9.33\widehat{i}+8.33\widehat{j}\right)\text{m/s}\end{array}$

Thus, the velocity of third particle is $\left(-9.33\times {10}^6\widehat{i}-8.33\times {10}^6\widehat{j}\right)\text{m/s}$.

(b)

To determine

The total kinetic energy increase in the process.

Answer to Problem 43P

The total kinetic energy increase in the process is $4.39\times {10}^{-13}\text{J}$ .

Explanation of Solution

Initially nucleus is at rest hence increase in kinetic energy is the final kinetic energy.

    $\Delta K=K_f-K_i$

Here, $\Delta K$ is the increase in kinetic energy, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy.

Substitute $0\text{J}$ for $K_i$ and solve for $\Delta K$ in above equation.

    $\Delta K=K_f$                                                                                                    (III)

Write the expression for the final kinetic energy of first particle.

    $K_1=\frac{1}{2}{m}_1{v}_1^2$                                                                                                (IV)

Here, $K_1$ is the kinetic energy of first particle.

Write the expression for the final kinetic energyof second particle .

    $K_2=\frac{1}{2}{m}_2{v}_2^2$                                                                                                 (V)

Here, $K_2$ is the kinetic energy of second particle.

Write the expression for the final kinetic energyof third particle .

    $K_3=\frac{1}{2}{m}_3{v}_3^2$                                                                                                (VI)

Here, $K_3$ is the kinetic energy of third particle.

    $K_f=K_1+K_2+K_3$                                                                                    (VII)

Substitute $\frac{1}{2}{m}_1{v}_1^2$ for $K_1$ , $\frac{1}{2}{m}_2{v}_2^2$ for $K_2$ and $\frac{1}{2}{m}_3{v}_3^2$ for $K_3$ in above equation.

    $K_f=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2+\frac{1}{2}{m}_3{v}_3^2$

Substitute $\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2+\frac{1}{2}{m}_3{v}_3^2$ for $K_f$ in equation (III) and solve for $\Delta K$.

    $\Delta K=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2+\frac{1}{2}{m}_3{v}_3^2$                                                                (VIII)

Conclusion:

Substitute $5\times {10}^{-27}\text{kg}$ for $m_1$, $\left(6\times {10}^6\text{m/s}\right)\widehat{j}$ for $v_1$, $8.4\times {10}^{-27}\text{kg}$ for $m_2$, $\left(4\times {10}^6\text{m/s}\right)\widehat{i}$ for $v_2$, $3.6\times {10}^{-27}\text{kg}$ for $m_3$ and  $\left(-9.33\times {10}^6\widehat{i}-8.33\times {10}^6\widehat{j}\right)\text{m/s}$ for $v_3$ in equation (VIII).

    $\begin{array}{c}\Delta K=\left[\begin{array}{l}\frac{1}{2}\left(5\times {10}^{-27}\text{kg}\right){\left(\left(6\times {10}^6\text{m/s}\right)\widehat{j}\right)}^2+\frac{1}{2}\left(8.4\times {10}^{-27}\text{kg}\right){\left(\left(4\times {10}^6\text{m/s}\right)\widehat{i}\right)}^2\\ +\frac{1}{2}\left(3.6\times {10}^{-27}\text{kg}\right)\left({\left(-9.33\times {10}^6\widehat{i}\right)}^2+{\left(-8.33\times {10}^6\widehat{j}\text{m/s}\right)}^2\right)\end{array}\right]\\ =\left(9\times {10}^{-14}\text{ J}\right)+\left(6.72\times {10}^{-14}\text{ J}\right)+\left(2.816\times {10}^{-13}\right)\\ =4.39\times {10}^{-13}\text{J}\end{array}$

Thus, the total kinetic energy increase in the process is $4.39\times {10}^{-13}\text{J}$ .